Mind Your Sum

Algebra Level 4

S = 2 3 − 5 6 + 8 12 − 11 24 + ⋯ + ( − 1 ) n 3 n + 2 3 × 2 n + ⋯ \large S=\dfrac{2}{3}-\dfrac{5}{6}+\dfrac{8 }{12}-\dfrac{11}{24}+\cdots + (-1)^n \frac{ 3n+2} { 3 \times 2^n } + \cdots

If S = a b 2 S = \dfrac{a}{b^2} , where a a and b b are positive coprime integers, find b ! a ! \dfrac{b!}{a!} .

Notation: ! ! denotes the factorial notation . For example: 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1 \times 2 \times 3 \times \cdots \times 8 .


The answer is 3.

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Satwik Murarka by Satwik Murarka , 20, India

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1 solution

S = 2 3 − 5 6 + 2 3 − 11 24 + ⋯ = 2 3 − 5 6 + 8 12 − 11 24 + ⋯ = 3 ( 0 ) + 2 3 ⋅ 2 0 − 3 ( 1 ) + 2 3 ⋅ 2 1 + 3 ( 2 ) + 2 3 ⋅ 2 2 − 3 ( 3 ) + 2 3 ⋅ 2 3 + ⋯ = ∑ n = 0 ∞ ( − 1 ) n ( 3 n + 2 ) 3 ⋅ 2 n = ∑ n = 0 ∞ ( − 1 ) n n 2 n + 2 3 ∑ n = 0 ∞ ( − 1 ) n 2 n = S 1 + 2 3 S 2 \begin{aligned} S & = \frac 23 - \frac 56 + {\color{#3D99F6}\frac 23} - \frac {11}{24} + \cdots \\ & = \frac 23 - \frac 56 + {\color{#3D99F6}\frac {8}{12}} - \frac {11}{24} + \cdots \\ & = \frac {3(0)+2}{3\cdot 2^0} - \frac {3(1)+2}{3\cdot 2^1} + \frac {3(2)+2}{3\cdot 2^2} - \frac {3(3)+2}{3\cdot 2^3} + \cdots \\ & = \sum_{n=0}^\infty \frac {(-1)^n(3n+2)}{3\cdot 2^n} \\ & = \sum_{n=0}^\infty \frac {(-1)^n n}{2^n} + \frac 23 \sum_{n=0}^\infty \frac {(-1)^n}{2^n} \\ & = S_1 + \frac 23 S_2 \end{aligned}

First consider S 2 S_2 :

S 2 = ∑ n = 0 ∞ ( − 1 ) n 2 n = 1 − 1 2 + 1 4 − 1 8 + ⋯ 2 S 2 = 2 − 1 + 1 2 − 1 4 + 1 8 − ⋯ = 2 − S 2 3 S 2 = 2 \begin{aligned} S_2 & = \sum_{n=0}^\infty \frac {(-1)^n}{2^n} \\ & = 1 - \frac 12 + \frac 14 - \frac 18 + \cdots \\ 2S_2 & = 2 - 1 + \frac 12 - \frac 14 + \frac 18 - \cdots \\ & = 2 - S_2 \\ 3S_2 & = 2 \end{aligned}

⟹ S 2 = 2 3 \begin{aligned} \implies S_2 & = \frac 23 \end{aligned}

Now S 1 S_1 :

S 1 = ∑ n = 0 ∞ ( − 1 ) n n 2 n = − 1 2 + 2 4 − 3 8 + 4 16 ⋯ 2 S 1 = − 1 + 2 2 − 3 4 + 4 8 ⋯ 3 S 2 = − 1 + 2 − 1 2 − 3 − 2 4 + 4 − 3 8 − ⋯ 3 S 2 = − 1 + 1 2 − 1 4 + 1 8 − ⋯ 3 S 2 = − S 2 = − 2 3 \begin{aligned} S_1 & = \sum_{n=0}^\infty \frac {(-1)^nn}{2^n} \\ & = - \frac 12 + \frac 24 - \frac 38 + \frac 4{16} \cdots \\ 2S_1 & = - 1 + \frac 22 - \frac 34 + \frac 48 \cdots \\ 3S_2 & = - 1+ \frac {2-1}2 - \frac {3-2}4 + \frac {4-3}8 - \cdots \\ 3S_2 & = - 1 + \frac 12 - \frac 14 + \frac 18 - \cdots \\ 3S_2 & = - S_2 = - \frac 23 \end{aligned}

⟹ S 1 = − 2 9 \begin{aligned} \implies S_1 & = - \frac 29 \end{aligned}

Therefore, S = S 1 + 2 3 S 2 = − 2 9 + 2 3 ⋅ 2 3 = 2 9 = 2 3 2 S = S_1 + \dfrac 23 S_2 = - \dfrac 29 + \dfrac 23 \cdot \dfrac 23 = \dfrac 29 = \dfrac 2{3^2}

⟹ b ! a ! = 3 ! 2 ! = 3 \implies \dfrac {b!}{a!} = \dfrac {3!}{2!} = \boxed{3}

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how did you get that generalization in the 3rd step?? can you please elaborate?? thank you.

mohan nayak - 4 years ago

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I have added a step to show. You should notice that the nominators have a common difference of 3 and the denominator, a common ratio of 2. That is how you get the formula.

Chew-Seong Cheong - 4 years ago

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oh, thanks.

mohan nayak - 4 years ago

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